# Rolling, torque and kinetic energy

1. Oct 6, 2014

### Lord Anoobis

1. The problem statement, all variables and given/known data
A uniform wheel of mass 10.0kg and radius 0.400m is mounted rigidly on a massless axle through its centre. The radius of the axle is 0.200m, and the rotational inertia of the wheel-axle combination about its central axis is 0.600kg.m2. The wheel is initially at rest at the top of a surface that is inclined at an angle of 30.0o with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by 2.00m, what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

2. Relevant equations

3. The attempt at a solution
The COM of the wheel drops a distance of h = mgsin30o = 1.00m, so the gain in kinetic energy is:

$$\Delta$$K = mgh = 10.0(9.80)(1.00) = 98.0J

and $$\Delta$$K = 0.5mV2 + 0.5I$$\omega$$2

So this is where things became sticky. I took the force due to gravity to be acting at the COM, and with the radius of the axle I arrived at a torque of mgsin30o(0.200) = 9.8N.m. Dividing this by the given rotational inertia I get $$\alpha$$ = 16.3rad/s2.

Next, the axle has a circumference of 0.4$$\pi$$, and with the 2.00m travelled I get an angular displacement of 10rad. In the end, plugging these values into the usual energy equations do not yield the correct answer for (a). I suspect I made a balls of it somewhere with the torque. Sorry about the rotten format folks. As my name indicates , I am but a noob here and I haven't gotten the hang of LaTex yet.

2. Oct 6, 2014

### Simon Bridge

I don't think you need to use torques - as a straight energy question: the loss of gravitational PE goes to rotating the wheel and also to translating it's center of mass. But the rotation and the translation are related because of the fact the rolling is without slipping.
It can help to draw a sketch.

Note: best practise - do not put the values in to the equation until after the algebra is done.
I'll help out with the LaTeX, here:

$mgh = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

OK so if the radius of the wheel is R, and the radius of the axel is r, then you can find a relation for $I$ and $\omega$ in terms of m, and R, and r, and v?

Last edited: Oct 6, 2014
3. Oct 6, 2014

### Lord Anoobis

Sorry, I don't quite see what you mean. Do you mean another relation in order to solve simultaneously or something else entirely?

4. Oct 6, 2014

### haruspex

Simon is saying there is a geometric relationship between the linear speed, the angular speed, and one or both of the given radii. Once you have that you can use your energy equation to deduce the angular and linear speeds.

5. Oct 6, 2014

### Lord Anoobis

Got it. Substitute the inner radius times ω for v and there it is. So painfully obvious I could kick myself. Thanks. Still, using the force of gravity as a torque could work as well yes? Where did I go wrong there?

6. Oct 6, 2014

### BvU

You forgot that the torque $\tau= mgr\;\sin\phi$ as you calculate it, is around a point on the circumference of the axle, not around the axle center. $I$ is different then (but that's not enough for me to get a match, so someone will have to jump on this to set us right..)

 made a small mistake. Using $I = 0.6 + mr^2 = {3\over 8}mR^2 + m{R\over 2}^2$ you get $\alpha = 9.8$ rad/s2 which gives you the right t for 2 m and the right $\omega$. All is well.

Last edited: Oct 6, 2014
7. Oct 6, 2014

### haruspex

A rolling wheel actually rotates about its point of contact with the ground. You measured the torque with respect to that point, but not the moment of inertia.

8. Oct 6, 2014

### Simon Bridge

You don't even need to calculate the speeds either - you are asked only for the energies.
Since you have clearly computed the solution now it is safe for me to do this:

$U=K+K_{rot} : U=mgh,\; K=\frac{1}{2}mv^2$... find Krot in terms of K.
$$K_{rot}=\frac{1}{2}I\omega^2 = \frac{1}{2} \left(\frac{1}{2}mR^2\right)\frac{v^2}{r^2}\\ \qquad = \frac{1}{2}mv^2 \frac{R^2}{2r^2}\\ \qquad = \frac{R^2}{2r^2}K$$
Substitute and solve for the energy you need directly without having to mess about with speeds. (This is one advantage of doing all the algebra before putting the numbers in.)

I think there is a deeper lesson here though - sometimes a problem requires more than one or two steps to solve: you also need to keep in mind the relationships between the various physical quantities. In this case the clue is that one part of the equation uses linear terms and the other uses rotational terms - if you make both parts use the same terms you can see the patterns more easily.

Last edited: Oct 6, 2014
9. Oct 13, 2014

### Lord Anoobis

Sorry for the late response, folks. I can't help feeling a bit chuffed that the remaining problems at the end of the chapter have not presented any difficulties or confusion at all after your inputs. Thanks a bunch once again.

10. Oct 13, 2014

### dean barry

If you consider an example constant velocity for the rig (say 10 m/s), then figure the KE linear and KE rotating, you can fix the ratio of the two, then from the translated energy you can figure the answers.

11. Oct 13, 2014

### dean barry

At example 10 m/s, the linear KE = ½ * 10 * 10 ² = 500 Joules
At 10 m/s the rotation rate = v / r = 10 / 0.2 = 50 rad / sec
The rotational KE at linear 10 m/s = ½ * 0.600 * 50 ² = 750 Joules

So, the KE ratio ( linear : rotating ) = 500 : 750 = 1 : 1.5

The portion of translated energy captured by rotation = 98 * ( 1.5 / ( 1.5 + 1 ) ) = 58.8 Joules